A scooter of 40,kg mass moving with velocity | vedclass

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(C) Initial kinetic energy $K_i = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} (40)(4)^2 + \frac{1}{2} (60)(2)^2 = 320 + 120 = 440\,J$.

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(C) Initial kinetic energy $K_i = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} (40)(4)^2 + \frac{1}{2} (60)(2)^2 = 320 + 120 = 440\,J$.By the Law of Conservation of Linear Momentum,$m_1 v_1 + m_2 v_2 = (m_1 + m_2) V_c$.$40(4) + 60(2) = (40 + 60) V_c \implies 160 + 120 = 100 V_c \implies 280 = 100 V_c \implies V_c = 2.8\,m/s$.Final kinetic energy $K_f = \frac{1}{2} (m_1 + m_2) V_c^2 = \frac{1}{2} (100) (2.8)^2 = 50 	imes 7.84 = 392\,J$.Loss in kinetic energy $\Delta K = K_i - K_f = 440 - 392 = 48\,J$.Alternatively,for a perfectly inelastic collision: $(\Delta K)_{	ext{loss}} = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (v_1 - v_2)^2 = \frac{1}{2} 	imes \frac{40 	imes 60}{100} 	imes (4 - 2)^2 = \frac{1}{2} 	imes 24 	imes 4 = 48\,J$.

$A$ scooter of $40\,kg$ mass moving with velocity $4\,m/s$ collides with another scooter of $60\,kg$ mass moving with velocity $2\,m/s$ in the same direction. After the collision,the two scooters stick to each other. The loss in kinetic energy is .............. $J$.

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